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Fiesta28 [93]
3 years ago
8

What is the score range for the smallest percent of students

Mathematics
1 answer:
andre [41]3 years ago
8 0
<h2>Answer:</h2>

The score range for the smallest percent of students is:

                         400 to 700

<h2>Step-by-step explanation:</h2>

From the pie chart or circle graph that is provided to us the smallest percent of the students are 10%

( Since 10%  is smallest among 14%,42%,34% and 10%)

and we may observe that the smallest region is denoted  or represented by the dark blue region and the range of the score of blue region that is given to us is:

                  400 to 700

   Hence, the answer is:

                   400 to 700

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You would multiply 4 on both sides
5x=2
then you would divide 5 on both sides
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A store offers four different brands of a product. It decides to eliminate the
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Brand D has the highest return rate so they should eliminate that one.

Step-by-step explanation:

brand A - .0481

brand B - .0307

Brand C - .0410

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3 years ago
The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.
salantis [7]

a) X

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

\mu=78.1 (mean)

\sigma = 10.8 (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

X

And the probability for this to occur can be written as:

p(X

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between z=-\infty and z=Z, where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

p(X

And by looking at the z-score tables, we find that this probability is:

p(z

And so,

p(X

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

p(X>89.1)

First of all, we have to calculate the z-score corresponding to this value of X, which is:

Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

p(z>1.02) =p(z

Because the normal distribution is symmetric.

But from part b) we know that

p(z

Therefore:

p(X>89.1)=p(z>1.02)=0.1539

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

p(67.1

Or also as

p(67.1

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

p(X

p(X>89.1)=0.1539

Therefore, here we find immediately than:

p(67.1

7 0
3 years ago
Solve the system of equations using the substitution method.y=2,2x+y=8
OlgaM077 [116]

Answer:

x=3

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
A set of middle school student heights are normally distributed with a mean of 150 centimeters and a standard deviation of 20 ce
julsineya [31]

Answer:

The proportion of students whose height are lower than Darnell's height is 71.57%

Step-by-step explanation:

The complete question is:

A set of middle school student heights are normally distributed with a mean of 150 centimeters and a standard deviation of 20 centimeters. Darnel is a middle school student with a height of 161.4cm.

What proportion of proportion of students height are lower than Darnell's height.

Answer:

We first calculate the z-score corresponding to Darnell's height using:

Z=\frac{X-\mu}{\sigma}

We substitute x=161.4 , \mu=150, and \sigma=20 to get:

Z=\frac{161.4-150}{20} \\Z=0.57

From the normal distribution table, we read 0.5 under 7.

The corresponding area is 0.7157

Therefore the proportion of students whose height are lower than Darnell's height is 71.57%

8 0
3 years ago
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